Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ : \(x\ne\pm\frac{1}{2}\)
\(E=\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}-\frac{\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\left(\frac{\left(1+2x\right)\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}-\frac{\left(1-2x\right)\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\)
\(E=\left(\frac{16x^4+8x^3+4x^2+2x+16x^4-8x^3-4x^2+2x}{1-16x^4}\right):\left(\frac{1+2x+x^2-1+2x-x^2}{1-4x^2}\right)\)
\(E=\frac{32x^4+4x}{1-16x^4}:\frac{4x}{1-4x^2}\)
\(E=\frac{4x\left(8x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{4x}\)
\(E=\frac{8x^3+1}{1+4x^2}\)
Study well
E=\(\left(\frac{4x^2+2x}{1-4x^2}-\frac{4x^2-2x}{1+4x^2}\right):\left(\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}\right)\)
E=\(\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\)\(\left(\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-\left(4x^2-16x^4-2x+8x^3\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{\left(1+4x+4x^2\right)-\left(1-4x+4x^2\right)}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^4+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{1+4x+4x^2-1+4x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{16x^4+2x+16x^4+2x}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{8x}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{32x^4+8x}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{8x\left(4x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{4x^3+1}{1+4x^2}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{1}{2}\\x\ne-\frac{1}{2}\\x\ne0\end{matrix}\right.\)
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\left[1:\left(1-\frac{1}{x}+\frac{1}{4x^2}\right)\right]\)
\(=\left[\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right]:\left[1:\frac{4x^2-4x+1}{4x^2}\right]\)
\(=\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x^2}{\left(2x-1\right)^2}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{\left(2x-1\right)^2}{4x^2}=\frac{2\left(2x-1\right)}{\left(2x+1\right).x}=\frac{4x-2}{2x^2+x}\left(ĐPCM\right)\)
\(A=\left(\frac{3x}{1-2x}-\frac{2x}{1+2x}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\frac{3x+2x}{1-2x}:\frac{2x^2+5}{\left(1-2x\right)^2}\)
\(A=\frac{5x}{1-2x}\cdot\frac{\left(1-2x\right)^2}{2x^2+5}\)
\(A=\frac{5x\left(1-2x\right)\left(1-2x\right)}{\left(1-2x\right)\left(2x^2+5\right)}\)
\(A=\frac{5x\left(1-2x\right)}{2x^2+5}\)
\(A=\frac{5x-10x^2}{2x^2+5}\)
\(A=\left(\frac{3x}{1-2x}-\frac{2x}{1+2x}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x.\left(1+2x\right)}{\left(1-2x\right).\left(1+2x\right)}-\frac{2x.\left(1-2x\right)}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x+6x^2}{\left(1-2x\right).\left(1+2x\right)}-\frac{2x-4x^2}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x+6x^2}{\left(1-2x\right).\left(1+2x\right)}+\frac{-\left(2x-4x^2\right)}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x+6x^2-2x+4x^2}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\frac{x+10x^2}{\left(1-2x\right).\left(1+2x\right)}:\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\frac{x+10x^2}{\left(1-2x\right).\left(1+2x\right)}:\frac{2x^2+5}{\left(1-2x\right)^2}\)
\(A=\frac{x+10x^2}{\left(1-2x\right).\left(1+2x\right)}.\frac{\left(1-2x\right)^2}{2x^2+5}\)
\(A=\frac{\left(x+10x^2\right).\left(1-2x\right)^2}{\left(1-2x\right).\left(1+2x\right).\left(2x^2+5\right)}\)
\(A=\frac{\left(x+10x^2\right).\left(1-2x\right)}{\left(1+2x\right).\left(2x^2+5\right)}\)
\(A=\frac{x-2x^2+10x^2-20x^3}{2x^2+5+4x^3+10x}\)
\(A=\frac{x+8x^2-20x^3}{2x^2+5+4x^3+10x}\)
Chúc bạn học tốt!
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)
\(=\frac{1}{ab}\)
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+14xy+y^2}{16x}\)
\(=\frac{\left(2x+y\right)^2+2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{x}{\left(2x-y\right)^2}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
ĐK: a, b khác 0, a khác -b
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{a+b}{ab}\right)\right].\frac{ab}{\left(a+b\right)^2}\)
\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}\right].\frac{ab}{\left(a+b\right)^2}=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)
\(A=\frac{\left(a+b\right)^2}{ab}.\frac{ab}{\left(a+b\right)^2}=1\)
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(4x^2-y^2\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16xy}\)
ĐK: xy khác 0, y \(\ne\pm\)2x
\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(2x-y\right).\left(2x+y\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\left[\frac{1}{\left(2x-y\right)}+\frac{1}{\left(2x+y\right)}\right]^2.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\left(\frac{2x+y+2x-y}{\left(2x-y\right).\left(2x+y\right)}\right)^2.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\frac{16x^2}{\left(2x-y\right)^2.\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16xy}\)
\(B=\frac{x}{\left(2x-y\right)^2.y}\)
E=\(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{1-4x^2}\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^2+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{1+4x+4x^2-1+4x-4x^2}\)
E=\(\frac{32x^4+4x}{8x\left(1+4x^2\right)}=\frac{8x^3+1}{2\left(1+4x^2\right)}\)
Mơn~